SOLUTION MANUAL
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS Q Q Q
1 - Nim and Combinatorial Games
Q Q Q Q Q
2 - Congestion Games
Q Q Q
3 - Games in Strategic Form
Q Q Q Q Q
4 - Game Trees with Perfect Information
Q Q Q Q Q Q
5 - Expected Utility
Q Q Q
6 - Mixed Equilibrium
Q Q Q
7 - Brouwer’s Fixed-Point Theorem
Q Q Q Q
8 - Zero-Sum Games
Q Q Q
9 - Geometry of Equilibria in Bimatrix Games
Q Q Q Q Q Q Q
10 - Game Trees with Imperfect Information
Q Q Q Q Q Q
11 - Bargaining
Q Q
12 - Correlated Equilibrium
Q Q Q
2
,Game Theory Basics
Q Q
SolutionsQ toQ Exercises
©Q BernhardQvonQStengelQ2022
SolutionQtoQExerciseQ1.1
(a) LetQ≤QbeQdefinedQbyQ(1.7).Q ToQshowQthatQ≤QisQtransitive,QconsiderQx,Qy,QzQwithQxQ ≤QyQandQyQ≤Q
z.QIfQxQ=QyQthenQxQ≤Qz,QandQifQyQ=QzQthenQalsoQxQ≤Qz.QSoQtheQonlyQcaseQleftQisQxQ<QyQandQ y
Q <Q z,QwhichQimpliesQ xQ <Q zQbecauseQ<QisQtransitive,QandQhenceQ xQ ≤Qz.
Clearly,Q≤QisQreflexiveQbecauseQxQ=QxQandQthereforeQxQ≤Qx.
ToQshowQthatQQQQQ
≤ isQantisymmetric,QconsiderQxQandQyQwithQxQQQQQ
≤ yQandQyQQQQQ
≤ x.QIfQweQha
dQxQ≠QyQthenQxQ<QyQandQyQ<Qx,QandQbyQtransitivityQxQ<QxQwhichQcontradictsQ(1.38).QHenceQ
xQ=Q y,QasQrequired.Q ThisQshowsQthatQ≤QisQaQpartialQorder.
Finally,QweQshowQ(1.6),QsoQweQhaveQtoQshowQthatQxQ<QyQimpliesQxQQQyQand
≤ QxQ≠QyQandQviceQ
versa.QLetQxQ<Qy,QwhichQimpliesQxQyQbyQ(1.7).≤QIfQweQhadQxQ=QyQthenQxQ<Qx,QcontradictingQ(1.
38),QsoQweQalsoQhaveQxQ≠Qy.Q Conversely,QxQQQ yQandQxQ≠QyQimply
≤QbyQ(1.7)QxQ <Q yQorQxQ =Q yQwhe
reQtheQsecondQcaseQisQexcluded,QhenceQ xQ <Q y,QasQrequired.
(b) ConsiderQaQpartialQorderQand ≤ QassumeQ(1.6)QasQaQdefinitionQofQ<.QToQshowQthatQ<QisQtransi
tive,QsupposeQxQ<Qy,QthatQis,QxQyQandQx≤
Q≠Qy,QandQyQ<Qz,QthatQis,QyQzQandQyQ≠Qz.QBecauseQQQQi
≤
sQtransitive,
≤ QxQQQQz.QIfQweQhadQxQ=QzQthenQxQQQQQyQandQyQQQQQxQandQhenceQxQ=QyQbyQantisy
≤ ≤ ≤
mmetryQofQQQQ,QwhichQcontradictsQ xQ ≠Q y,QsoQweQhaveQ xQQQQzQandQ xQ ≠Q z,QthatQis,QxQ <Q zQbyQ(
≤ ≤
1.6),QasQrequired.
Also,Q<QisQirreflexive,QbecauseQxQ<QxQwouldQbyQdefinitionQmeanQxQQQxQand
≤ QxQ≠Qx,QbutQtheQl
atterQisQnotQtrue.
Finally,QweQshowQ(1.7),QsoQweQhaveQtoQshowQthatQxQ ≤QyQimpliesQxQ<QyQorQxQ=QyQandQviceQv
ersa,QgivenQthatQ<QisQdefinedQbyQ(1.6).QLetQxQ≤Qy.QThenQifQxQ=Qy,QweQareQdone,QotherwiseQx
Q≠QyQandQthenQbyQdefinitionQxQ<Qy.QHence,QxQ≤QyQimpliesQxQ<QyQorQxQ=Qy.QConversely,Qsupp
oseQxQ <Q yQorQxQ=Qy.Q IfQxQ <Q yQthenQxQ ≤QyQbyQ(1.6),QandQifQxQ=QyQthenQxQ ≤Q yQbecauseQ≤Qis
Qreflexive.Q ThisQcompletesQtheQproof.
SolutionQtoQExerciseQ1.2
(a) InQ analysingQ theQ gamesQ ofQ threeQ NimQ heapsQ whereQ oneQ heapQ hasQ sizeQ one,Q weQ firstQ lookQatQs
omeQexamples,QandQthenQuseQmathematicalQinductionQtoQproveQwhatQweQconjectureQtoQbeQtheQlo
singQpositions.QAQlosingQpositionQisQoneQwhereQeveryQmoveQisQtoQaQwinningQposition,Qbeca
useQthenQtheQopponentQwillQwin.Q TheQpointQofQthisQexerciseQisQtoQformulateQaQpreciseQstatem
entQtoQbeQproved,QandQthenQtoQproveQit.
First,QifQthereQareQonlyQtwoQheapsQrecallQthatQtheyQareQlosingQifQandQonlyQifQtheQheapsQare
QofQequalQsize.Q IfQtheyQareQofQunequalQsize,QthenQtheQwinningQmoveQisQtoQreduceQtheQlarger
QheapQsoQthatQbothQheapsQhaveQequalQsize.
3
, ConsiderQthreeQheapsQofQsizesQ1,Qm,Qn,QwhereQ1QQQQQ ≤ mQQQQQ
≤ n.QWeQobserveQtheQfollowing:Q
1,Q1,QmQisQwinning,QbyQmovingQtoQ1,Q1,Q0.QSimilarly,Q1,Qm,QmQisQwinning,QbyQmovingQtoQ0,
Qm,Qm.QNext,Q1,Q2,Q3QisQlosingQ(observedQearlierQinQtheQlecture),QandQhenceQ1,Q2,QnQforQnQ
4QisQwinning.Q1,Q3,QnQisQwinningQforQanyQnQ3QbyQmovingQtoQ1,Q3,Q2.QForQ1,Q4,Q5,Qreducing
≥ ≥
QanyQheapQproducesQaQwinningQposition,QsoQthisQisQlosing.
TheQgeneralQpatternQforQtheQlosingQpositionsQthusQseemsQtoQbe:Q1,Qm,QmQ1,Qfor
+ QevenQnum
bersQm.Q ThisQincludesQalsoQtheQcaseQmQ=Q0,QwhichQweQcanQtakeQasQtheQbaseQcaseQforQanQind
uction.Q WeQnowQproceedQtoQproveQthisQformally.
FirstQweQshowQthatQifQtheQpositionsQofQtheQformQ1,Qm,QnQwithQmQQQQQQ ≤ nQareQlosingQwhenQ
mQisQevenQandQnQ=QmQ1, +QthenQtheseQareQtheQonlyQlosingQpositionsQbecauseQanyQotherQpositi
onQ1,Qm,QnQ withQmQ Q nQ is≤
Qwinning.Q Namely,QifQmQ =QnQ thenQaQwinningQmoveQfromQ1,Qm,QmQi
sQtoQ0,Qm,Qm,QsoQweQcanQassumeQmQ<Qn.Q IfQmQisQevenQthenQnQ>QmQ Q 1Q(otherwiseQweQwouldQb
+
eQinQtheQpositionQ1,Qm,QmQ Q 1)QandQsoQtheQwinningQmoveQisQtoQ1,Qm,QmQ Q 1.QIfQmQisQoddQth
+ +
enQtheQwinningQmoveQisQtoQ1,Qm,QmQ1,QtheQsameQasQpositionQ1,QmQ1,QmQ(thisQwouldQ alsoQ beQ aQ
winningQ moveQ fromQ 1,Qm,QmQ soQ thereQ theQ winning–Q moveQ isQ notQ unique). −
Second,QweQshowQthatQanyQmoveQfromQ1,Qm,QmQ+Q1QwithQevenQmQisQtoQaQwinningQposition,Qusi
ngQasQinductiveQhypothesisQthatQ1,QmJ,QmJQ+Q1QforQevenQmJQandQmJQ<QmQisQaQlosingQposition
.QTheQmoveQtoQ0,Qm,QmQ+Q1QproducesQaQwinningQpositionQwithQcounter-
moveQtoQ0,Qm,Qm.QAQmoveQtoQ1,QmJ,QmQ+Q1QforQmJQ<QmQisQtoQaQwinningQpositionQwithQtheQcou
nter-
moveQtoQ1,QmJ,QmJQ+Q1QifQmJQisQevenQandQtoQ1,QmJ,QmJQ−Q1QifQmJQisQodd.QAQmoveQtoQ1,Qm,Q
mQisQtoQaQwinningQpositionQwithQcounter-
moveQtoQ0,Qm,Qm.QAQmoveQtoQ1,Qm,QmJQwithQ mJQ<Q mQisQalsoQtoQaQwinningQpositionQwithQtheQc
ounter-
moveQtoQ1,QmJQ−Q1,Qm+J J J J J
+QevenQ(inQwhichQcaseQmJQ 1Q
QifQ m QisQodd,QandQtoQ1,Qm Q 1,Qm QifQm Qis
<QmQbecauseQmQisQeven).QThisQconcludesQtheQinductionQproof.
ThisQresultQisQinQagreementQwithQtheQtheoremQonQNimQheapQsizesQrepresentedQasQsumsQofQpow
ersQofQ2:Q 1Q Q∗Qm+∗ 0
Q Q nQisQlosingQifQandQonlyQif,QexceptQforQ2 ,QtheQpowersQofQ2QmakingQupQmQan
+∗
dQnQcomeQinQpairs.QSoQtheseQmustQbeQtheQsameQpowersQofQ2,QexceptQforQ1Q=Q20,QwhichQoccurs
QinQonlyQmQorQn,QwhereQweQhaveQassumedQthatQnQisQtheQlargerQnumber,QsoQ1QappearsQinQ theQ re
presentationQ ofQ n:Q WeQ haveQ mQ =Q 2aQQQQQQ2bQQQQQQ2c
+ + +Q ·Q ·Q ·
forQ aQ >Q bQ >Q cQ >QQQQQQQQ
·Qa·+
QQ ·Q
Q≥Q +bQ Q Q+Q ·cQ ·Q ·Q + +
1,QsoQ mQ isQ even,Q and,Q withQ theQ sameQ a,Qb,Qc,Q.Q.Q.,Q nQ =Q 2 2 2
1Q =Q mQQQQ 1.Q Then
∗1QQQQQQQ
+Q ∗ + mQQQQQQ
∗ nQQQQQQ 0.Q TheQ followingQ isQ anQ exampleQ usingQ theQ bitQ representationQ where
mQ=Q12Q(which
≡Q∗ QdeterminesQtheQbitQpatternQ1100,QwhichQofQcourseQdependsQonQm):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) WeQuseQ(a).QClearly,Q1,Q2,Q3QisQlosingQasQshownQinQ(1.2),QandQbecauseQtheQNim-
sumQofQtheQbinaryQrepresentationsQ01,Q10,Q11QisQ00.QExamplesQshowQthatQanyQotherQposit
ionQisQwinning.QTheQthreeQnumbersQareQn,+QnQ 1,Q nQ Q 2.QIfQnQisQevenQthenQreducingQtheQheapQ
+
ofQsizeQnQ2QtoQ1QcreatesQtheQpositionQn,QnQ 1,Q1QwhichQisQlosingQasQshownQinQ(a).QIfQnQisQo
+ +
dd,QthenQnQ 1QisQevenQandQnQQQ2Q=Q nQQQ1QQQ1QsoQbyQtheQsameQargument,QaQwinningQmov
+ +
eQisQtoQreduceQtheQNimQheapQofQsizeQnQtoQ1Q(whichQonlyQworksQifQnQ >Q1).
(Q +Q )Q+
4
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS Q Q Q
1 - Nim and Combinatorial Games
Q Q Q Q Q
2 - Congestion Games
Q Q Q
3 - Games in Strategic Form
Q Q Q Q Q
4 - Game Trees with Perfect Information
Q Q Q Q Q Q
5 - Expected Utility
Q Q Q
6 - Mixed Equilibrium
Q Q Q
7 - Brouwer’s Fixed-Point Theorem
Q Q Q Q
8 - Zero-Sum Games
Q Q Q
9 - Geometry of Equilibria in Bimatrix Games
Q Q Q Q Q Q Q
10 - Game Trees with Imperfect Information
Q Q Q Q Q Q
11 - Bargaining
Q Q
12 - Correlated Equilibrium
Q Q Q
2
,Game Theory Basics
Q Q
SolutionsQ toQ Exercises
©Q BernhardQvonQStengelQ2022
SolutionQtoQExerciseQ1.1
(a) LetQ≤QbeQdefinedQbyQ(1.7).Q ToQshowQthatQ≤QisQtransitive,QconsiderQx,Qy,QzQwithQxQ ≤QyQandQyQ≤Q
z.QIfQxQ=QyQthenQxQ≤Qz,QandQifQyQ=QzQthenQalsoQxQ≤Qz.QSoQtheQonlyQcaseQleftQisQxQ<QyQandQ y
Q <Q z,QwhichQimpliesQ xQ <Q zQbecauseQ<QisQtransitive,QandQhenceQ xQ ≤Qz.
Clearly,Q≤QisQreflexiveQbecauseQxQ=QxQandQthereforeQxQ≤Qx.
ToQshowQthatQQQQQ
≤ isQantisymmetric,QconsiderQxQandQyQwithQxQQQQQ
≤ yQandQyQQQQQ
≤ x.QIfQweQha
dQxQ≠QyQthenQxQ<QyQandQyQ<Qx,QandQbyQtransitivityQxQ<QxQwhichQcontradictsQ(1.38).QHenceQ
xQ=Q y,QasQrequired.Q ThisQshowsQthatQ≤QisQaQpartialQorder.
Finally,QweQshowQ(1.6),QsoQweQhaveQtoQshowQthatQxQ<QyQimpliesQxQQQyQand
≤ QxQ≠QyQandQviceQ
versa.QLetQxQ<Qy,QwhichQimpliesQxQyQbyQ(1.7).≤QIfQweQhadQxQ=QyQthenQxQ<Qx,QcontradictingQ(1.
38),QsoQweQalsoQhaveQxQ≠Qy.Q Conversely,QxQQQ yQandQxQ≠QyQimply
≤QbyQ(1.7)QxQ <Q yQorQxQ =Q yQwhe
reQtheQsecondQcaseQisQexcluded,QhenceQ xQ <Q y,QasQrequired.
(b) ConsiderQaQpartialQorderQand ≤ QassumeQ(1.6)QasQaQdefinitionQofQ<.QToQshowQthatQ<QisQtransi
tive,QsupposeQxQ<Qy,QthatQis,QxQyQandQx≤
Q≠Qy,QandQyQ<Qz,QthatQis,QyQzQandQyQ≠Qz.QBecauseQQQQi
≤
sQtransitive,
≤ QxQQQQz.QIfQweQhadQxQ=QzQthenQxQQQQQyQandQyQQQQQxQandQhenceQxQ=QyQbyQantisy
≤ ≤ ≤
mmetryQofQQQQ,QwhichQcontradictsQ xQ ≠Q y,QsoQweQhaveQ xQQQQzQandQ xQ ≠Q z,QthatQis,QxQ <Q zQbyQ(
≤ ≤
1.6),QasQrequired.
Also,Q<QisQirreflexive,QbecauseQxQ<QxQwouldQbyQdefinitionQmeanQxQQQxQand
≤ QxQ≠Qx,QbutQtheQl
atterQisQnotQtrue.
Finally,QweQshowQ(1.7),QsoQweQhaveQtoQshowQthatQxQ ≤QyQimpliesQxQ<QyQorQxQ=QyQandQviceQv
ersa,QgivenQthatQ<QisQdefinedQbyQ(1.6).QLetQxQ≤Qy.QThenQifQxQ=Qy,QweQareQdone,QotherwiseQx
Q≠QyQandQthenQbyQdefinitionQxQ<Qy.QHence,QxQ≤QyQimpliesQxQ<QyQorQxQ=Qy.QConversely,Qsupp
oseQxQ <Q yQorQxQ=Qy.Q IfQxQ <Q yQthenQxQ ≤QyQbyQ(1.6),QandQifQxQ=QyQthenQxQ ≤Q yQbecauseQ≤Qis
Qreflexive.Q ThisQcompletesQtheQproof.
SolutionQtoQExerciseQ1.2
(a) InQ analysingQ theQ gamesQ ofQ threeQ NimQ heapsQ whereQ oneQ heapQ hasQ sizeQ one,Q weQ firstQ lookQatQs
omeQexamples,QandQthenQuseQmathematicalQinductionQtoQproveQwhatQweQconjectureQtoQbeQtheQlo
singQpositions.QAQlosingQpositionQisQoneQwhereQeveryQmoveQisQtoQaQwinningQposition,Qbeca
useQthenQtheQopponentQwillQwin.Q TheQpointQofQthisQexerciseQisQtoQformulateQaQpreciseQstatem
entQtoQbeQproved,QandQthenQtoQproveQit.
First,QifQthereQareQonlyQtwoQheapsQrecallQthatQtheyQareQlosingQifQandQonlyQifQtheQheapsQare
QofQequalQsize.Q IfQtheyQareQofQunequalQsize,QthenQtheQwinningQmoveQisQtoQreduceQtheQlarger
QheapQsoQthatQbothQheapsQhaveQequalQsize.
3
, ConsiderQthreeQheapsQofQsizesQ1,Qm,Qn,QwhereQ1QQQQQ ≤ mQQQQQ
≤ n.QWeQobserveQtheQfollowing:Q
1,Q1,QmQisQwinning,QbyQmovingQtoQ1,Q1,Q0.QSimilarly,Q1,Qm,QmQisQwinning,QbyQmovingQtoQ0,
Qm,Qm.QNext,Q1,Q2,Q3QisQlosingQ(observedQearlierQinQtheQlecture),QandQhenceQ1,Q2,QnQforQnQ
4QisQwinning.Q1,Q3,QnQisQwinningQforQanyQnQ3QbyQmovingQtoQ1,Q3,Q2.QForQ1,Q4,Q5,Qreducing
≥ ≥
QanyQheapQproducesQaQwinningQposition,QsoQthisQisQlosing.
TheQgeneralQpatternQforQtheQlosingQpositionsQthusQseemsQtoQbe:Q1,Qm,QmQ1,Qfor
+ QevenQnum
bersQm.Q ThisQincludesQalsoQtheQcaseQmQ=Q0,QwhichQweQcanQtakeQasQtheQbaseQcaseQforQanQind
uction.Q WeQnowQproceedQtoQproveQthisQformally.
FirstQweQshowQthatQifQtheQpositionsQofQtheQformQ1,Qm,QnQwithQmQQQQQQ ≤ nQareQlosingQwhenQ
mQisQevenQandQnQ=QmQ1, +QthenQtheseQareQtheQonlyQlosingQpositionsQbecauseQanyQotherQpositi
onQ1,Qm,QnQ withQmQ Q nQ is≤
Qwinning.Q Namely,QifQmQ =QnQ thenQaQwinningQmoveQfromQ1,Qm,QmQi
sQtoQ0,Qm,Qm,QsoQweQcanQassumeQmQ<Qn.Q IfQmQisQevenQthenQnQ>QmQ Q 1Q(otherwiseQweQwouldQb
+
eQinQtheQpositionQ1,Qm,QmQ Q 1)QandQsoQtheQwinningQmoveQisQtoQ1,Qm,QmQ Q 1.QIfQmQisQoddQth
+ +
enQtheQwinningQmoveQisQtoQ1,Qm,QmQ1,QtheQsameQasQpositionQ1,QmQ1,QmQ(thisQwouldQ alsoQ beQ aQ
winningQ moveQ fromQ 1,Qm,QmQ soQ thereQ theQ winning–Q moveQ isQ notQ unique). −
Second,QweQshowQthatQanyQmoveQfromQ1,Qm,QmQ+Q1QwithQevenQmQisQtoQaQwinningQposition,Qusi
ngQasQinductiveQhypothesisQthatQ1,QmJ,QmJQ+Q1QforQevenQmJQandQmJQ<QmQisQaQlosingQposition
.QTheQmoveQtoQ0,Qm,QmQ+Q1QproducesQaQwinningQpositionQwithQcounter-
moveQtoQ0,Qm,Qm.QAQmoveQtoQ1,QmJ,QmQ+Q1QforQmJQ<QmQisQtoQaQwinningQpositionQwithQtheQcou
nter-
moveQtoQ1,QmJ,QmJQ+Q1QifQmJQisQevenQandQtoQ1,QmJ,QmJQ−Q1QifQmJQisQodd.QAQmoveQtoQ1,Qm,Q
mQisQtoQaQwinningQpositionQwithQcounter-
moveQtoQ0,Qm,Qm.QAQmoveQtoQ1,Qm,QmJQwithQ mJQ<Q mQisQalsoQtoQaQwinningQpositionQwithQtheQc
ounter-
moveQtoQ1,QmJQ−Q1,Qm+J J J J J
+QevenQ(inQwhichQcaseQmJQ 1Q
QifQ m QisQodd,QandQtoQ1,Qm Q 1,Qm QifQm Qis
<QmQbecauseQmQisQeven).QThisQconcludesQtheQinductionQproof.
ThisQresultQisQinQagreementQwithQtheQtheoremQonQNimQheapQsizesQrepresentedQasQsumsQofQpow
ersQofQ2:Q 1Q Q∗Qm+∗ 0
Q Q nQisQlosingQifQandQonlyQif,QexceptQforQ2 ,QtheQpowersQofQ2QmakingQupQmQan
+∗
dQnQcomeQinQpairs.QSoQtheseQmustQbeQtheQsameQpowersQofQ2,QexceptQforQ1Q=Q20,QwhichQoccurs
QinQonlyQmQorQn,QwhereQweQhaveQassumedQthatQnQisQtheQlargerQnumber,QsoQ1QappearsQinQ theQ re
presentationQ ofQ n:Q WeQ haveQ mQ =Q 2aQQQQQQ2bQQQQQQ2c
+ + +Q ·Q ·Q ·
forQ aQ >Q bQ >Q cQ >QQQQQQQQ
·Qa·+
QQ ·Q
Q≥Q +bQ Q Q+Q ·cQ ·Q ·Q + +
1,QsoQ mQ isQ even,Q and,Q withQ theQ sameQ a,Qb,Qc,Q.Q.Q.,Q nQ =Q 2 2 2
1Q =Q mQQQQ 1.Q Then
∗1QQQQQQQ
+Q ∗ + mQQQQQQ
∗ nQQQQQQ 0.Q TheQ followingQ isQ anQ exampleQ usingQ theQ bitQ representationQ where
mQ=Q12Q(which
≡Q∗ QdeterminesQtheQbitQpatternQ1100,QwhichQofQcourseQdependsQonQm):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) WeQuseQ(a).QClearly,Q1,Q2,Q3QisQlosingQasQshownQinQ(1.2),QandQbecauseQtheQNim-
sumQofQtheQbinaryQrepresentationsQ01,Q10,Q11QisQ00.QExamplesQshowQthatQanyQotherQposit
ionQisQwinning.QTheQthreeQnumbersQareQn,+QnQ 1,Q nQ Q 2.QIfQnQisQevenQthenQreducingQtheQheapQ
+
ofQsizeQnQ2QtoQ1QcreatesQtheQpositionQn,QnQ 1,Q1QwhichQisQlosingQasQshownQinQ(a).QIfQnQisQo
+ +
dd,QthenQnQ 1QisQevenQandQnQQQ2Q=Q nQQQ1QQQ1QsoQbyQtheQsameQargument,QaQwinningQmov
+ +
eQisQtoQreduceQtheQNimQheapQofQsizeQnQtoQ1Q(whichQonlyQworksQifQnQ >Q1).
(Q +Q )Q+
4
